Introduction to Open Data Science - Course Project

About the project

Write a short description about the course and add a link to your GitHub repository here. This is an R Markdown (.Rmd) file so you should use R Markdown syntax.

# This is a so-called "R chunk" where you can write R code.

date()
## [1] "Thu Nov 19 11:22:05 2020"
"I am Ambrin from the University of Eastern Finland"
## [1] "I am Ambrin from the University of Eastern Finland"
"I am based in Kuopio."
## [1] "I am based in Kuopio."
"I am doing my PhD in Health Sciences"
## [1] "I am doing my PhD in Health Sciences"
"I am planning to learn basics of R and concepts about open data science here"
## [1] "I am planning to learn basics of R and concepts about open data science here"
"I am curious"
## [1] "I am curious"
"I heard about this course from the yammer platform of my University"
## [1] "I heard about this course from the yammer platform of my University"
"Its great to learn new things."
## [1] "Its great to learn new things."
"The link to my github repository is https://github.com/ambrinbabu/IODS-project/"
## [1] "The link to my github repository is https://github.com/ambrinbabu/IODS-project/"

The text continues here.

:/home/Ambrin$ git config –global Ambrin.email


Insert chapter 2 title here

Describe the work you have done this week and summarize your learning.

date()
## [1] "Thu Nov 19 11:22:05 2020"

We would like to read the students2014 data into R from the url provided and would like to explore the data.

#Read the students2014 data into R from url and store in students2014
students2014 <- read.table("http://s3.amazonaws.com/assets.datacamp.com/production/course_2218/datasets/learning2014.txt",sep=",", header=TRUE )
head(students2014)
##   gender age attitude     deep  stra     surf points
## 1      F  53      3.7 3.583333 3.375 2.583333     25
## 2      M  55      3.1 2.916667 2.750 3.166667     12
## 3      F  49      2.5 3.500000 3.625 2.250000     24
## 4      M  53      3.5 3.500000 3.125 2.250000     10
## 5      M  49      3.7 3.666667 3.625 2.833333     22
## 6      F  38      3.8 4.750000 3.625 2.416667     21
#Explore the structure of the data 
str(students2014)
## 'data.frame':    166 obs. of  7 variables:
##  $ gender  : chr  "F" "M" "F" "M" ...
##  $ age     : int  53 55 49 53 49 38 50 37 37 42 ...
##  $ attitude: num  3.7 3.1 2.5 3.5 3.7 3.8 3.5 2.9 3.8 2.1 ...
##  $ deep    : num  3.58 2.92 3.5 3.5 3.67 ...
##  $ stra    : num  3.38 2.75 3.62 3.12 3.62 ...
##  $ surf    : num  2.58 3.17 2.25 2.25 2.83 ...
##  $ points  : int  25 12 24 10 22 21 21 31 24 26 ...

The data is in the form of a data frame and as we can see, there are 7 columns of data representing gender, age, attitude, deep, stra, surf and points. There are 166 observations (rows). The data comes from ASSIST (The approaches and study skills inventory for students) and includes information about 116 students of different age groups and comprise both male and female and approaches they use for learning - surface approach (memorise without understanding with a serious lack of personal engagement in learning process), deep approach (intention to maximise understanding with a true commitment to learning) and strategic approach (apply any strategy that maximises the chance of chieving highest possible grades). The student achievments are measured by points in the exams

#Explore the dimensions of the data
dim(students2014)
## [1] 166   7
#ggplot2 is a popular library for creating stunning graphics with R.
#install.packages("ggplot2")
library(ggplot2)
#Show a graphical overview of the data 
p1 <- ggplot(students2014, aes(x = attitude, y = points, col = gender))

# define the visualization type (points)
p2 <- p1 + geom_point()


# add a regression line
p3 <- p2 + geom_smooth(method = "lm")


# add a main title and draw the plot
p4 <- p3+ ggtitle("Student's attitude versus exam points")
p4
## `geom_smooth()` using formula 'y ~ x'

Here we see a graphical overview of the data and see the relationship between the attitude vs exam points

#Show summaries of the variables in the data. Describe and interpret the outputs, commenting on the distributions of the variables and the relationships between them
##Simple regression

# a scatter plot of points versus attitude
library(ggplot2)
qplot(attitude, points, data = students2014) + geom_smooth(method = "lm")
## `geom_smooth()` using formula 'y ~ x'

# fit a linear model
my_model <- lm(points ~ attitude, data = students2014)

# print out a summary of the model

summary(my_model)
## 
## Call:
## lm(formula = points ~ attitude, data = students2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -16.9763  -3.2119   0.4339   4.1534  10.6645 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  11.6372     1.8303   6.358 1.95e-09 ***
## attitude      3.5255     0.5674   6.214 4.12e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared:  0.1906, Adjusted R-squared:  0.1856 
## F-statistic: 38.61 on 1 and 164 DF,  p-value: 4.119e-09

We see here a linear model fitted for our data.Linear regression model is a simple statistical model.It is an approach for modelling the relationship between a dependent variable y and one or more exploratory variables x.The model is found by minimising the sum of the residuals.Residuals are essentially the difference between the actual observed response values and the response values that the model predicted.

The exam points are the target variable and attitude is the explanatory variable. The variables are evenly distributed. The summary of the variables in the data is also shown.The Residuals section of the model output breaks it down into 5 summary points - min, 1Q, median, 2Q and max. The coefficients gives estimates for the parameters of the model.In this case the p value for attitude is very low. So there is a statistical relationship between attitude and points.

#Choose three variables as explanatory variables and fit a regression model where exam points is the target (dependent) variable. Show a summary of the fitted model and comment and interpret the results. Explain and interpret the statistical test related to the model parameters. If an explanatory variable in your model does not have a statistically significant relationship with the target variable, remove the variable from the model and fit the model again without it.
my_model2 <- lm(points ~ attitude + stra + surf, data = students2014)
summary(my_model2)
## 
## Call:
## lm(formula = points ~ attitude + stra + surf, data = students2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -17.1550  -3.4346   0.5156   3.6401  10.8952 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  11.0171     3.6837   2.991  0.00322 ** 
## attitude      3.3952     0.5741   5.913 1.93e-08 ***
## stra          0.8531     0.5416   1.575  0.11716    
## surf         -0.5861     0.8014  -0.731  0.46563    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared:  0.2074, Adjusted R-squared:  0.1927 
## F-statistic: 14.13 on 3 and 162 DF,  p-value: 3.156e-08

Here 3 variables (attitude, strategic learning, surface learning) are explanatory variables and a multiple regression model is fitted where points is the target variable. There are 5 point summary of residuals of model - min, 1Q, median, 3Q and Max.

Below the residuals section, we see coefficients which gives estimates for the parameters of the model. The estimate corresponding to the intercept is the estimate of alpha parameter and the estimate corresponding to attitude is beta parameter. Here we estimated the effect of attitude on points to be 11.01 with standard difference of approx 3.68. We also have t and p values corresponding to statistics test of null hypothesis that the actual value of beta parameter would be 0. In this case the p value for attitude is very low. So there is a statistical relationship between attitude and points.However, for stra and surf, these values are not statistically significant and so there is no statistical relationship between stra (or surf) and points.

Residual Standard Error is the measure of the quality of a linear regression fit. Theoretically, every linear model is assumed to contain an error term E. Due to the presence of this error term, we are not capable of perfectly predicting our explanatory variable from the target variable. The Residual Standard Error is the average amount that the response will deviate from the true regression line. It’s also worth noting that the Residual Standard Error was calculated with 162 degrees of freedom. Simplistically, degrees of freedom are the number of data points that went into the estimation of the parameters used after taking into account these parameters.

The R-squared (R2) statistic provides a measure of how well the model is fitting the actual data. It takes the form of a proportion of variance. R2 is a measure of the linear relationship between our target variable and our explanatory variable. It always lies between 0 and 1 (i.e.: a number near 0 represents a regression that does not explain the variance in the response variable well and a number close to 1 does explain the observed variance in the response variable). In our example, the R2 we get is 0.2074. Or roughly 20% of the variance found in the explanatory variables can be explained by the target variable.

#If an explanatory variable in your model does not have a statistically significant relationship with the target variable, remove the variable from the model and fit the model again without it.
my_model2 <- lm(points ~ attitude, data = students2014)
summary(my_model2)
## 
## Call:
## lm(formula = points ~ attitude, data = students2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -16.9763  -3.2119   0.4339   4.1534  10.6645 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  11.6372     1.8303   6.358 1.95e-09 ***
## attitude      3.5255     0.5674   6.214 4.12e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared:  0.1906, Adjusted R-squared:  0.1856 
## F-statistic: 38.61 on 1 and 164 DF,  p-value: 4.119e-09

In the previous example, we saw that 2 of the explanatory variables (strategic and surface learning) did not have a statistically significant relationship with the target variable (points). Hence we removed the 2 variables and fitted the model again without it.

#Produce the following diagnostic plots: Residuals vs Fitted values, Normal QQ-plot and Residuals vs Leverage
# create a regression model with multiple explanatory variables
my_model2 <- lm(points ~ attitude + stra, data = students2014)
# draw diagnostic plots using the plot() function. Choose the plots 1, 2 and 5
#1 residuals vs fitted values
#2 Normal QQplot
#5 Residuals vs levarage
par(mfrow = c(2,2))
plot(my_model2, which = 2)

Statistical models always include several assumption which describe the data generating process. In a linear regression model, we assume linearity. The target variable is modelled as a linear combination of the model parameters. Usually it is assumed that the errors are normally distributed, not correlated and have constant variance. Further, its also assumed that the size of a given error does not depend on the values of the explanatory variables.

QQ-plot: QQ plot of the residuals provides a meathod to explore the assumption that the errors of the model are normally distributed. The better the points fall within the line, the better is the fit to the normality assumption. In our case, we see a reasonable fit.

#Produce the following diagnostic plots: Residuals vs Fitted values, Normal QQ-plot and Residuals vs Leverage
# create a regression model with multiple explanatory variables
my_model2 <- lm(points ~ attitude + stra, data = students2014)
# draw diagnostic plots using the plot() function. Choose the plots 1, 2 and 5
#1 residuals vs fitted values
#2 Normal QQplot
#5 Residuals vs levarage
par(mfrow = c(2,2))
plot(my_model2, which = 1)

The constant variance assumption implies that the size of the errors should not depend on the explanatory variables.This can be explored by plotting a scatter plot of residuals versus model predictors. In our case, we dont see when fitted values increase, spread of residuals increase, indicating a problem.

#Produce the following diagnostic plots: Residuals vs Fitted values, Normal QQ-plot and Residuals vs Leverage
# create a regression model with multiple explanatory variables
my_model2 <- lm(points ~ attitude + stra, data = students2014)
# draw diagnostic plots using the plot() function. Choose the plots 1, 2 and 5
#1 residuals vs fitted values
#2 Normal QQplot
#5 Residuals vs levarage
par(mfrow = c(2,2))
plot(my_model2, which = 5)

Leverage of observations measures how much impact a single observation has on the model.Residuals vs leverage plot can help identify which observations have an unusually high impact.We do not have one particular point with very high leverage so we can conclude that it is a regular leverage without any outliers.


(more chapters to be added similarly as we proceed with the course!)

Chapter 3

Describe the work you have done this week and summarize your learning.

date()
## [1] "Thu Nov 19 11:22:09 2020"
# read data 
alc <- read.csv("https://github.com/rsund/IODS-project/raw/master/data/alc.csv")
# explore structure and dimension
str(alc)
## 'data.frame':    370 obs. of  51 variables:
##  $ school    : chr  "GP" "GP" "GP" "GP" ...
##  $ sex       : chr  "F" "F" "F" "F" ...
##  $ age       : int  15 15 15 15 15 15 15 15 15 15 ...
##  $ address   : chr  "R" "R" "R" "R" ...
##  $ famsize   : chr  "GT3" "GT3" "GT3" "GT3" ...
##  $ Pstatus   : chr  "T" "T" "T" "T" ...
##  $ Medu      : int  1 1 2 2 3 3 3 2 3 3 ...
##  $ Fedu      : int  1 1 2 4 3 4 4 2 1 3 ...
##  $ Mjob      : chr  "at_home" "other" "at_home" "services" ...
##  $ Fjob      : chr  "other" "other" "other" "health" ...
##  $ reason    : chr  "home" "reputation" "reputation" "course" ...
##  $ guardian  : chr  "mother" "mother" "mother" "mother" ...
##  $ traveltime: int  2 1 1 1 2 1 2 2 2 1 ...
##  $ studytime : int  4 2 1 3 3 3 3 2 4 4 ...
##  $ schoolsup : chr  "yes" "yes" "yes" "yes" ...
##  $ famsup    : chr  "yes" "yes" "yes" "yes" ...
##  $ activities: chr  "yes" "no" "yes" "yes" ...
##  $ nursery   : chr  "yes" "no" "yes" "yes" ...
##  $ higher    : chr  "yes" "yes" "yes" "yes" ...
##  $ internet  : chr  "yes" "yes" "no" "yes" ...
##  $ romantic  : chr  "no" "yes" "no" "no" ...
##  $ famrel    : int  3 3 4 4 4 4 4 4 4 4 ...
##  $ freetime  : int  1 3 3 3 2 3 2 1 4 3 ...
##  $ goout     : int  2 4 1 2 1 2 2 3 2 3 ...
##  $ Dalc      : int  1 2 1 1 2 1 2 1 2 1 ...
##  $ Walc      : int  1 4 1 1 3 1 2 3 3 1 ...
##  $ health    : int  1 5 2 5 3 5 5 4 3 4 ...
##  $ n         : int  2 2 2 2 2 2 2 2 2 2 ...
##  $ id.p      : int  1096 1073 1040 1025 1166 1039 1131 1069 1070 1106 ...
##  $ id.m      : int  2096 2073 2040 2025 2153 2039 2131 2069 2070 2106 ...
##  $ failures  : int  0 1 0 0 1 0 1 0 0 0 ...
##  $ paid      : chr  "yes" "no" "no" "no" ...
##  $ absences  : int  3 2 8 2 5 2 0 1 9 10 ...
##  $ G1        : int  10 10 14 10 12 12 11 10 16 10 ...
##  $ G2        : int  12 8 13 10 12 12 6 10 16 10 ...
##  $ G3        : int  12 8 12 9 12 12 6 10 16 10 ...
##  $ failures.p: int  0 0 0 0 0 0 0 0 0 0 ...
##  $ paid.p    : chr  "yes" "no" "no" "no" ...
##  $ absences.p: int  4 2 8 2 2 2 0 0 6 10 ...
##  $ G1.p      : int  13 13 14 10 13 11 10 11 15 10 ...
##  $ G2.p      : int  13 11 13 11 13 12 11 10 15 10 ...
##  $ G3.p      : int  13 11 12 10 13 12 12 11 15 10 ...
##  $ failures.m: int  1 2 0 0 2 0 2 0 0 0 ...
##  $ paid.m    : chr  "yes" "no" "yes" "yes" ...
##  $ absences.m: int  2 2 8 2 8 2 0 2 12 10 ...
##  $ G1.m      : int  7 8 14 10 10 12 12 8 16 10 ...
##  $ G2.m      : int  10 6 13 9 10 12 0 9 16 11 ...
##  $ G3.m      : int  10 5 13 8 10 11 0 8 16 11 ...
##  $ alc_use   : num  1 3 1 1 2.5 1 2 2 2.5 1 ...
##  $ high_use  : logi  FALSE TRUE FALSE FALSE TRUE FALSE ...
##  $ cid       : int  3001 3002 3003 3004 3005 3006 3007 3008 3009 3010 ...
dim(alc)
## [1] 370  51

It is a dataframe with 370 observation and 51 variables. The data is about the secondary school in 2 portugese schools and their achievements. The data attributes include student grades, demographic, social and school related features) and it was collected by using school reports and questionnaires. Two datasets are provided regarding the performance in two distinct subjects: Mathematics (mat) and Portuguese language (por).The data contains information also about the different students, their age, sex etc. and if they consume high levels of alcohol or not. The alcohol consumption scale ranges from 1 (very low) to 5 (very high).

Hypothesis: My hypothesis is that many reasons affect the alcohol consumption.I have chosen 4 interesting variables - gender, famrel, goout (going out with friends) and freetime. Gender- I hypothesise that male drink more than women. famrel - When there are more family relations, less alcohol is consumed goout - when someone goes out with friends, more alcohol is consumed freetime - when there is more freetime, more alcohol is consumed

#Numerically and graphically explore the distributions of your chosen variables and their relationships with alcohol consumption (use for example cross-tabulations, bar plots and box plots). Comment on your findings and compare the results of your exploration to your previously stated hypotheses.
library(tidyr); library(dplyr); library(ggplot2)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
gather(alc) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar()

alc %>% group_by(sex, high_use) %>% summarise(count = n())
## `summarise()` regrouping output by 'sex' (override with `.groups` argument)
## # A tibble: 4 x 3
## # Groups:   sex [2]
##   sex   high_use count
##   <chr> <lgl>    <int>
## 1 F     FALSE      154
## 2 F     TRUE        41
## 3 M     FALSE      105
## 4 M     TRUE        70
alc %>% group_by(failures, high_use) %>% summarise(count = n())
## `summarise()` regrouping output by 'failures' (override with `.groups` argument)
## # A tibble: 8 x 3
## # Groups:   failures [4]
##   failures high_use count
##      <int> <lgl>    <int>
## 1        0 FALSE      238
## 2        0 TRUE        87
## 3        1 FALSE       12
## 4        1 TRUE        12
## 5        2 FALSE        8
## 6        2 TRUE         9
## 7        3 FALSE        1
## 8        3 TRUE         3
alc %>% group_by(famrel, high_use) %>% summarise(count = n())
## `summarise()` regrouping output by 'famrel' (override with `.groups` argument)
## # A tibble: 10 x 3
## # Groups:   famrel [5]
##    famrel high_use count
##     <int> <lgl>    <int>
##  1      1 FALSE        6
##  2      1 TRUE         2
##  3      2 FALSE        9
##  4      2 TRUE         9
##  5      3 FALSE       39
##  6      3 TRUE        25
##  7      4 FALSE      128
##  8      4 TRUE        52
##  9      5 FALSE       77
## 10      5 TRUE        23
alc %>% group_by(goout, high_use) %>% summarise(count = n())
## `summarise()` regrouping output by 'goout' (override with `.groups` argument)
## # A tibble: 10 x 3
## # Groups:   goout [5]
##    goout high_use count
##    <int> <lgl>    <int>
##  1     1 FALSE       19
##  2     1 TRUE         3
##  3     2 FALSE       82
##  4     2 TRUE        15
##  5     3 FALSE       97
##  6     3 TRUE        23
##  7     4 FALSE       40
##  8     4 TRUE        38
##  9     5 FALSE       21
## 10     5 TRUE        32

Gender - more male students drink more (according to hypothesis) Famrel - More relatives around, less alcohol is consumed (according to hypothesis) goout - When students go out, more alcohol is consumed (according to hypothesis) freetime - more free time, more students drink alcohol (according to hypothesis)

#Graphical results

g1<- ggplot(alc, aes(x = high_use, y = famrel))+ geom_boxplot() + ggtitle("Family relationship vs alcohol consumption")
g1

g2<-ggplot(alc, aes(x = high_use, y = famrel, col = sex))+ geom_boxplot() + ggtitle("Family relationship vs alcohol consumption according to gender")
g2

g3<- ggplot(alc, aes(x = high_use, y = goout))+ geom_boxplot() + ggtitle("Going out with friends vs alcohol consumption")
g3

g4<-ggplot(alc, aes(x = high_use, y = goout, col = sex))+ geom_boxplot() + ggtitle("Going out with friends vs alcohol consumption according to gender")
g4

g5<- ggplot(alc, aes(x = high_use, y = freetime))+ geom_boxplot() + ggtitle("Freetime vs alcohol consumption")
g5

g6<-ggplot(alc, aes(x = high_use, y = freetime, col = sex))+ geom_boxplot() + ggtitle("Freetime vs alcohol consumption according to gender")
g6

Logistic regression

m <- glm(high_use ~ goout + famrel + sex+ freetime, data = alc, family = "binomial")
summary(m)
## 
## Call:
## glm(formula = high_use ~ goout + famrel + sex + freetime, family = "binomial", 
##     data = alc)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.6569  -0.7847  -0.5274   0.8571   2.5801  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  -2.4926     0.6994  -3.564 0.000365 ***
## goout         0.7702     0.1267   6.078 1.22e-09 ***
## famrel       -0.4483     0.1414  -3.171 0.001519 ** 
## sexM          0.9558     0.2599   3.677 0.000236 ***
## freetime      0.1120     0.1400   0.800 0.423896    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 452.04  on 369  degrees of freedom
## Residual deviance: 378.47  on 365  degrees of freedom
## AIC: 388.47
## 
## Number of Fisher Scoring iterations: 4

We see here a linear model fitted for our data.Linear regression model is a simple statistical model.It is an approach for modelling the relationship between a dependent variable y and one or more exploratory variables x.The model is found by minimising the sum of the residuals.Residuals are essentially the difference between the actual observed response values and the response values that the model predicted.

The high alcohol use is the target variable and goout + famrel + sex+ freetime is the explanatory variable. The variables are evenly distributed. The summary of the variables in the data is also shown.The Residuals section of the model output breaks it down into 5 summary points - min, 1Q, median, 2Q and max. The coefficients gives estimates for the parameters of the model.In this case the p value for goout, famrel, and sex is very low. So there is a statistical relationship between going out with friends, family relationships and sex with high alcohol use.Free time has p value above 0.05 so there is no statistical relationship between high alcohol use and free time.

#Present and interpret the coefficients of the model as odds ratios and provide confidence intervals for them. Interpret the results and compare them to your previously stated hypothesis.
coef(m)
## (Intercept)       goout      famrel        sexM    freetime 
##  -2.4926487   0.7701903  -0.4483457   0.9557531   0.1119551
OR <- coef(m) %>% exp
CI <- confint(m) %>% exp 
## Waiting for profiling to be done...
cbind(OR, CI)
##                     OR      2.5 %    97.5 %
## (Intercept) 0.08269065 0.02016869 0.3154115
## goout       2.16017719 1.69707698 2.7921949
## famrel      0.63868384 0.48200915 0.8405944
## sexM        2.60062833 1.57104372 4.3613169
## freetime    1.11846266 0.85030850 1.4743895

Odd ratios s a statistic that quantifies the strength of the association between two events. If it is greater than 1 we have a positive association and if the odd ratio is smaller than 1 its a negative association. Going out with friends, sex and free time have a positve association with high alcohol use an family relation has negative association. The results are according to the stated hypothesis. The confidence intervals are widest for the the sex variable, so its effect is the most uncertain

Predictive model Predictive power of the final logistic regression model is calculated without the statistically insignificant variable - freetime

m <- glm(high_use ~ goout + famrel + sex, data = alc, family = "binomial")

# predict() the probability of high_use
probabilities <- predict(m, type = "response")

# add the predicted probabilities to 'alc'
alc <- mutate(alc, probability = probabilities)

# use the probabilities to make a prediction of high_use
alc <- mutate(alc, prediction = probability > 0.5)

select(alc, goout, famrel, sex, high_use, probability, prediction) %>% tail(10)
##     goout famrel sex high_use probability prediction
## 361     3      5   M     TRUE  0.25406225      FALSE
## 362     3      4   M     TRUE  0.34478448      FALSE
## 363     1      4   M     TRUE  0.09640288      FALSE
## 364     4      3   M     TRUE  0.64356587       TRUE
## 365     2      3   M    FALSE  0.26797363      FALSE
## 366     3      4   M     TRUE  0.34478448      FALSE
## 367     2      4   M    FALSE  0.19155369      FALSE
## 368     4      4   M     TRUE  0.53888551       TRUE
## 369     4      5   M     TRUE  0.43065938      FALSE
## 370     2      4   M    FALSE  0.19155369      FALSE
# creating a confusion matrix with actual values
table(high_use = alc$high_use, prediction = alc$prediction)
##         prediction
## high_use FALSE TRUE
##    FALSE   241   18
##    TRUE     62   49
 # creating a confusion matrix with predicted values
table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table %>% addmargins 
##         prediction
## high_use      FALSE       TRUE        Sum
##    FALSE 0.65135135 0.04864865 0.70000000
##    TRUE  0.16756757 0.13243243 0.30000000
##    Sum   0.81891892 0.18108108 1.00000000
g <- ggplot(alc, aes(x = probability, y = high_use, col = prediction))+ geom_point() + ggtitle("logistic regression model")
g 

241 students dont consume high levels of alcohol. 18 are predicted wrongly (dont drink alcohol but is predicted to be drinking high levels of alcohol). 49 students drink alcohol and prediction is correct but 62 students drink alcohol but it is predicted that they dont.

Training error

loss_func <- function(class, prob) {
  n_wrong <- abs(class - prob) > 0.5
  mean(n_wrong)
}
# compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability) 
## [1] 0.2162162
library(boot)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = nrow(alc)) # K-fold cross-validation, cv=training data
# average number of wrong predictions in the cross validation
cv$delta[1]
## [1] 0.2216216

Training error can be calculated by adding false positives and false negatives. This is further confirmed by loss function. Here, we can see the total proportion of inaccurately classified individuals. The number is about 22% and is not very high. So our model is performing good but can be improved.

Bonus question

loss_func <- function(class, prob) {
  n_wrong <- abs(class - prob) > 0.5
  mean(n_wrong)
}
 # compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)
## [1] 0.2162162
library(boot)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10) # K-fold cross-validation, cv=training data
# average number of wrong predictions in the cross validation
cv$delta[1]
## [1] 0.2351351

This model has a better test set performance (0.22) compared to model in Datacample(0.26). 10-fold cross-validation gives good estimate of the actual predictive power of the model. Low value = good

R Markdown

In this exercise we use Boston data from MASS-library. This dataset contains information collected by the U.S Census Service concerning housing in the area of Boston Mass. Data includes 14 variables and 506 rows

# access the MASS package and load other libraries for later analysis
library(MASS)
## 
## Attaching package: 'MASS'
## The following object is masked from 'package:dplyr':
## 
##     select
library(corrplot)
## corrplot 0.84 loaded
library(dplyr)
library(plotly)
## 
## Attaching package: 'plotly'
## The following object is masked from 'package:MASS':
## 
##     select
## The following object is masked from 'package:ggplot2':
## 
##     last_plot
## The following object is masked from 'package:stats':
## 
##     filter
## The following object is masked from 'package:graphics':
## 
##     layout
# load the data
 data("Boston")
str(Boston)
## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
dim(Boston)
## [1] 506  14
#Show a graphical overview of the data and show summaries of the variables in the data. Describe and interpret the outputs, commenting on the distributions of the variables and the relationships between them
summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08205   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00
pairs(Boston)

There are some very interesting distributions fo variables in the plot matrix. Variable rad has high and low values so the plot shows that the values are concentrated on either side of the plot.

#Correlation matrix#

#calculating the correlation matrix and correlation plot
cor_matrix <- round(cor(Boston),digits = 2)
corrplot(cor_matrix, method="circle", type = "upper", cl.pos = "b", tl.pos = "d", tl.cex = 0.6)

Plotted correlation matrix shows that there is some high correlation between variables: * Correlation is quite clear between industrial areas (indus) and nitrogen oxides (nox). Industry adds pollution in the area. Industry seems to correlate also with variablrs like age, dis, ras and tax. * Nitrogen oxides (nox) correlations are very similar with industry (indus) * Crime rate (crim) seems to correlate with good accessibilitty to radial highways (rad) and value property (tax). * Old houses (age) and employment centers have also something common

#Scaled data# All the variables are numerical so we can use scale()-function to scale whole data set

#Standardize the dataset and print out summaries of the scaled data. How did the variables change? 
boston_scaled <- scale(Boston)
summary(boston_scaled)
##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv        
##  Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 3.5453   Max.   : 2.9865
class(boston_scaled)
## [1] "matrix" "array"
boston_scaled <- as.data.frame(boston_scaled)

Scaling the data makes variables look as if they are in the same range. Variables like black and tax were before scaling hundred fold compared to some other variables

#Create a categorical variable of the crime rate in the Boston dataset (from the scaled crime rate). Use the quantiles as the break points in the categorical variable. Drop the old crime rate variable from the dataset. 
#save the scaled crim as scaled_crim
scaled_crim <- boston_scaled$crim
#create a quantile vector of crim and print it
bins <- quantile(scaled_crim)
bins
##           0%          25%          50%          75%         100% 
## -0.419366929 -0.410563278 -0.390280295  0.007389247  9.924109610
#create a categorical variable 'crime'
crime <- cut(scaled_crim, breaks = bins, include.lowest = TRUE, label = c("low", "med_low", "med_high", "high"))
#look at the table of the new factor crime, do not change the actual variable "crime"
crime_tab <-table(crime)
crime_tab
## crime
##      low  med_low med_high     high 
##      127      126      126      127
#remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
summary(boston_scaled)
##        zn               indus              chas              nox         
##  Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723   Min.   :-1.4644  
##  1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723   1st Qu.:-0.9121  
##  Median :-0.48724   Median :-0.2109   Median :-0.2723   Median :-0.1441  
##  Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723   3rd Qu.: 0.5981  
##  Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648   Max.   : 2.7296  
##        rm               age               dis               rad         
##  Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658   Min.   :-0.9819  
##  1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049   1st Qu.:-0.6373  
##  Median :-0.1084   Median : 0.3171   Median :-0.2790   Median :-0.5225  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617   3rd Qu.: 1.6596  
##  Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566   Max.   : 1.6596  
##       tax             ptratio            black             lstat        
##  Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033   Min.   :-1.5296  
##  1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049   1st Qu.:-0.7986  
##  Median :-0.4642   Median : 0.2746   Median : 0.3808   Median :-0.1811  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332   3rd Qu.: 0.6024  
##  Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406   Max.   : 3.5453  
##       medv              crime    
##  Min.   :-1.9063   low     :127  
##  1st Qu.:-0.5989   med_low :126  
##  Median :-0.1449   med_high:126  
##  Mean   : 0.0000   high    :127  
##  3rd Qu.: 0.2683                 
##  Max.   : 2.9865

#Train and test set# Training set contains 80% of the data. 20% is in the test set

#Divide the dataset to train and test sets, so that 80% of the data belongs to the train set
# number of rows in the Boston dataset 
n <- nrow(boston_scaled)
# choose randomly 80% of the rows
ind <- sample(n,  size = n * 0.8)
ind
##   [1] 189 104 403 453 134 448 475 379 423 296 185  65 154 352  64 246 490  73
##  [19] 473 159 302  63 255 209 165   8  74 345 320 182 419 322 223 156 455 288
##  [37] 427 421 214 369 250 395 160 153 300 442  12 211  60 479  99  29  71 291
##  [55] 360  88  35 164  97 181 241 233 335  55  21 271 264 217  87 113 155 413
##  [73] 429 408 258 315 391  94 452  66 277 274 265 215 309   3  20 334 424 487
##  [91] 366  27 136 458  34 422  18 180 170 105 194  61 261 440 299 117 219  19
## [109]  70 239 504  46 494 284 451 118 464   5 262 439  22 347  83 206 370 353
## [127]  25 171 240 390 257 416 188 179 232 441 205 460 147  17 149  15 186 411
## [145] 176 469 409 245 145  84   7 447 476 319 491  40  31  28 406 459 129 471
## [163] 138  38 236 192 228 357 398 287 174 466 275 168 397 254 454 348 201 308
## [181] 230 428  23 497 294 269  26 196 286  92 400 418 495 349 295 272 114 132
## [199]  89  54 417 336 387 229 237 298 216 327  37  43 260  95 346 162 467 425
## [217] 394 125 210  59 207 119 310 472  24 124 140 474 478 195 499 465 107 354
## [235] 102 371 221 289 363 436 183 115 306 173 242  72 414 148  47 383 135 244
## [253]  82 314  58  44 151 281 449 333 341 238  78   9 169 365 381 404 377 368
## [271] 317  42 410 480 304 468 268 364 235 456  90 131 218 415 133 492  76 126
## [289] 243 463 502 331 202 187 450  77 361 405 111  85 412 273  51 110  75 190
## [307] 328 431 344 312 279 338 305 270  56 139 401 378 109 224 142 372 399 128
## [325] 251 146 420 122  39 248 143 443 489 367 222  67 457 137 200 388 330 482
## [343] 144 384 506 350 163  79  30 259 503  80 175 100 355 106 484  32 488 249
## [361] 193  10 501 324 150 393 380 434 462  16  48  96 323 321 303 339 483 177
## [379]  93 373 493 231 112 166 375 337 283 392 234  53 141   6  14  81 247 204
## [397]  13 161 386  11 121 227 359 326
# create train set
train <- boston_scaled[ind,]
# create test set 
test <- boston_scaled[-ind,]

#Fitting the Linear Discriminant Analysis# First the linear discriminant analysis (LDA) is fitted to the train set. The new categorical variable crime is the target variable and all the other variables of the dataset are predictor variables. After fitting we draw the LDA biplot with arrows

#Fit the linear discriminant analysis on the train set. Use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables. Draw the LDA (bi)plot.
#LDA = linear discriminant analysis
lda.fit <- lda(crime ~. , data = train)
#print the lda.fit object
lda.fit
## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2376238 0.2450495 0.2623762 0.2549505 
## 
## Group means:
##                  zn      indus        chas        nox          rm        age
## low       0.9013537 -0.8650714 -0.10828322 -0.8517176  0.42567188 -0.8716644
## med_low  -0.1056770 -0.2539862 -0.03371693 -0.5766104 -0.14007951 -0.3728070
## med_high -0.3885419  0.1669443  0.24766530  0.3804698  0.09691199  0.4208289
## high     -0.4872402  1.0170891 -0.04298342  1.0449929 -0.43871124  0.8060424
##                 dis        rad        tax    ptratio      black       lstat
## low       0.8267243 -0.6732207 -0.7543941 -0.3412866  0.3747636 -0.75622521
## med_low   0.3589258 -0.5387253 -0.4473719 -0.1017024  0.3499318 -0.12719618
## med_high -0.3489345 -0.4314738 -0.3148704 -0.2239220  0.1045227  0.01197604
## high     -0.8516330  1.6384176  1.5142626  0.7811136 -0.7229913  0.84898651
##                 medv
## low       0.53702433
## med_low  -0.02674077
## med_high  0.12855003
## high     -0.62628309
## 
## Coefficients of linear discriminants:
##                 LD1         LD2         LD3
## zn       0.08306703  0.77172618 -1.03906861
## indus    0.04748549 -0.09394423  0.34229580
## chas    -0.09598624 -0.08548403  0.07740603
## nox      0.37099951 -0.71287138 -1.41062398
## rm      -0.12347542 -0.13069335 -0.15168260
## age      0.18329175 -0.44710766 -0.17143956
## dis     -0.09559292 -0.31480080  0.22721641
## rad      3.47448075  0.95322317 -0.06120180
## tax     -0.05805092 -0.14051149  0.74269590
## ptratio  0.15330851  0.07261385 -0.56944170
## black   -0.12832555  0.05037934  0.22558652
## lstat    0.18654097 -0.13651966  0.38431656
## medv     0.19545008 -0.24254472 -0.25590207
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9518 0.0336 0.0145
#the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}
#target classes as numeric
classes <- as.numeric(train$crime)
classes
##   [1] 2 2 4 4 3 4 4 4 4 2 2 1 3 1 2 2 2 2 3 3 1 2 1 2 3 2 2 1 3 1 4 2 3 3 4 1 4
##  [38] 4 2 4 2 4 3 3 1 4 2 2 2 4 1 3 2 1 4 1 3 3 2 1 2 3 1 1 3 3 3 1 1 2 3 4 4 4
##  [75] 3 3 4 1 4 1 2 2 3 3 3 1 3 1 4 4 4 3 3 4 3 4 3 1 3 2 1 2 3 4 1 2 2 3 2 2 1
## [112] 2 2 1 4 2 4 1 3 4 3 1 1 2 4 1 3 3 2 4 1 4 1 1 3 4 1 4 3 3 3 3 1 4 1 4 4 2
## [149] 3 1 2 4 4 3 2 1 3 3 4 4 3 4 3 1 3 1 3 4 4 1 2 3 1 3 4 3 4 1 1 1 3 4 3 3 2
## [186] 3 3 1 1 1 4 4 3 1 1 2 2 3 1 1 4 1 4 3 3 2 2 3 2 2 3 1 1 3 4 4 4 2 3 2 2 2
## [223] 3 4 3 2 3 4 4 1 2 4 2 1 2 4 3 1 4 4 2 2 1 2 2 2 4 3 2 4 3 2 1 3 1 2 3 1 4
## [260] 1 1 3 2 2 3 3 4 4 4 4 3 2 4 4 2 4 3 4 3 4 1 3 1 4 3 2 2 2 2 4 1 1 1 1 4 2
## [297] 4 4 2 1 4 2 2 3 1 2 2 4 1 3 1 1 1 2 1 2 4 4 2 3 3 4 4 3 2 3 4 1 2 2 3 4 2
## [334] 4 3 1 4 3 1 4 1 4 4 4 1 1 3 1 3 3 1 2 2 1 1 2 3 3 4 2 2 2 2 3 3 4 4 4 4 3
## [371] 2 2 3 2 2 1 4 1 1 4 2 3 2 3 4 1 1 4 3 1 3 1 3 1 3 1 2 3 4 2 1 3 4 2
#plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 3)

#Predicting the classes#

#Save the crime categories from the test set and then remove the categorical crime variable from the test dataset. Then predict the classes with the LDA model on the test data. Cross tabulate the results with the crime categories from the test set. Comment on the results
#save the correct classes from test data
correct_classes <- test$crime
#remove the crime variable from test data
test <- dplyr::select(test, -crime)
#predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
#cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)
##           predicted
## correct    low med_low med_high high
##   low       20      11        0    0
##   med_low    6      11       10    0
##   med_high   0       3       15    2
##   high       0       0        0   24

Prediction were quite good. There was some errors in the middle of the range but classes low and especially high were good. Only one correct representative of high class was predicted to med_low class.

#Reload the Boston dataset and standardize the dataset (we did not do this in the Datacamp exercises, but you should scale the variables to get comparable distances). Calculate the distances between the observations. Run k-means algorithm on the dataset. Investigate what is the optimal number of clusters and run the algorithm again. Visualize the clusters (for example with the pairs() or ggpairs() functions, where the clusters are separated with colors) and interpret the results
#Loading and scaling Boston data
scaled_Boston <- scale(Boston)
summary(scaled_Boston)
##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv        
##  Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 3.5453   Max.   : 2.9865
#calculating the euclidean distance matrix between the observation
dist_eu <- dist(scaled_Boston)
#determining the max number of clusters
cluster_max <- 15
#calculate the total within sum of squares
#K-means might produce different results every time, because it randomly 
#assigns the initial cluster centers. The function set.seed() can be used to 
#deal with that.
set.seed(123)
twcss <- sapply(1:cluster_max, function(k){kmeans(dist_eu, k)$tot.withinss})
# visualize the results
plot(1:cluster_max, twcss, type='b')

One way to determine the number of clusters is to look how the total of within cluster sum of squares (WCSS) behaves when the number of clusters changes. WCSS was calculated from 1 to 15 clusters. The optimal number of clusters is when the total WCSS drops radically. It seems that in this case optimal number of clusters is two. However we are here to learn so I decided to analyse model with four clusters.

After determining the number of clusters I run the K-means alcorithm again

#k-means clustering
km <-kmeans(dist_eu, centers = 4)
# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)

It seems that when the data is divided to four clusters there is some clear differences in distriputions of several variables. Crim, zn, indus and blacks are variables were one can distinguish clear patterns between clusters. Some variables (rad & tax) show that sometimes 1 or 2 clusters make a clear distripution but observation of other two clusters are ambigious and there is no clear pattern to be regognised.

#BONUS: LDA using clusters as target classes#

#Perform k-means on the original Boston data with some reasonable number of clusters (> 2). Remember to standardize the dataset. Then perform LDA using the clusters as target classes. Include all the variables in the Boston data in the LDA model. Visualize the results with a biplot (include arrows representing the relationships of the original variables to the LDA solution). Interpret the results. Which variables are the most influencial linear separators for the clusters? 
#Loading and scaling Boston data
scaled_Boston <- scale(Boston)
scaled_Boston <- as.data.frame(scaled_Boston)
#colnames(scaled_Boston)
#calculating the euclidean distance matrix between the observation
dist_eu <- dist(scaled_Boston)
#k-means clustering
set.seed(123)
km <-kmeans(dist_eu, centers = 4)
cm <- as.data.frame(km$cluster)
#adding the clusters to the scaled dataset
scaled_Boston <- data.frame(scaled_Boston, clust = cm)
colnames(scaled_Boston)[15] <- "clust"
summary(scaled_Boston)
##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv             clust      
##  Min.   :-1.5296   Min.   :-1.9063   Min.   :1.000  
##  1st Qu.:-0.7986   1st Qu.:-0.5989   1st Qu.:2.000  
##  Median :-0.1811   Median :-0.1449   Median :3.000  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   :2.943  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683   3rd Qu.:4.000  
##  Max.   : 3.5453   Max.   : 2.9865   Max.   :4.000
#Original Boston dataset is now scaled and the result of K-means clustering is saved to the variable *clust*


#LDA = linear discriminant analysis
lda.fit.km <- lda(clust ~. , data = scaled_Boston)
#print the lda.fit object
lda.fit.km
## Call:
## lda(clust ~ ., data = scaled_Boston)
## 
## Prior probabilities of groups:
##         1         2         3         4 
## 0.1304348 0.2272727 0.2114625 0.4308300 
## 
## Group means:
##         crim         zn      indus       chas        nox         rm        age
## 1  1.4330759 -0.4872402  1.0689719  0.4435073  1.3439101 -0.7461469  0.8575386
## 2  0.2797949 -0.4872402  1.1892663 -0.2723291  0.8998296 -0.2770011  0.7716696
## 3 -0.3912182  1.2671159 -0.8754697  0.5739635 -0.7359091  0.9938426 -0.6949417
## 4 -0.3894453 -0.2173896 -0.5212959 -0.2723291 -0.5203495 -0.1157814 -0.3256000
##          dis        rad        tax     ptratio       black      lstat
## 1 -0.9620552  1.2941816  1.2970210  0.42015742 -1.65562038  1.1930953
## 2 -0.7723199  0.9006160  1.0311612  0.60093343 -0.01717546  0.6116223
## 3  0.7751031 -0.5965444 -0.6369476 -0.96586616  0.34190729 -0.8200275
## 4  0.3182404 -0.5741127 -0.6240070  0.02986213  0.34248644 -0.2813666
##          medv
## 1 -0.81904111
## 2 -0.54636549
## 3  1.11919598
## 4 -0.01314324
## 
## Coefficients of linear discriminants:
##                 LD1        LD2         LD3
## crim    -0.18113078  0.5012256  0.60535205
## zn      -0.43297497  1.0486194 -0.67406151
## indus   -1.37753200 -0.3016928 -1.07034034
## chas     0.04307937  0.7598229  0.22448239
## nox     -1.04674638  0.3861005  0.33268952
## rm       0.14912869  0.1510367 -0.67942589
## age      0.09897424 -0.0523110 -0.26285587
## dis     -0.13139210  0.1593367  0.03487882
## rad     -0.65824136 -0.5189795 -0.48145070
## tax     -0.28903561  0.5773959 -0.10350513
## ptratio -0.22236843 -0.1668597  0.09181715
## black    0.42730704 -0.5843973 -0.89869354
## lstat   -0.24320629  0.6197780  0.01119242
## medv    -0.21961575  0.9485829  0.17065360
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.7596 0.1768 0.0636
#the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}
#target classes as numeric
classes <- as.numeric(scaled_Boston$clust)
#classes
#plot the lda results
plot(lda.fit.km, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit.km, myscale = 3)

#Super-bonus# 3D plot where observations color is the crime classes of the train set

model_predictors <- dplyr::select(train, -crime)
#check the dimensions
#dim(model_predictors)
#dim(lda.fit$scaling)
#matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
#3d plot
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)
## Warning: `arrange_()` is deprecated as of dplyr 0.7.0.
## Please use `arrange()` instead.
## See vignette('programming') for more help
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.

3D plot where observations color is based on the K-means clusters

model_predictors <- dplyr::select(scaled_Boston, -clust)
#check the dimensions
#dim(model_predictors)
#dim(lda.fit.km$scaling)
#matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit.km$scaling
matrix_product <- as.data.frame(matrix_product)
#3D plot
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = scaled_Boston$clust)

Colors of the both plots is based to four classes. It seems that K-means plot shows the different clusters more clearly than the plot that is based on the crime classification.